Magnetic field at the centre of a coil of 100 turns and radius `2xx10^(-3)` carrying 1 A current is:

Options

A. `3.14xx10^(-10)T`
B. `3.14xx10^(-7)T`
C. `3.14xx10^(7)T`
D. `3.14xx10^(10)T`

Explanation

To find the magnetic field at the center of a coil, we can use the formula:

\[ B = \frac{\mu_0 n I}{2R} \]

where:
- \(B\) is the magnetic field at the center of the coil,
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(n\) is the number of turns per unit length (in this case, total number of turns),
- \(I\) is the current in amperes,
- \(R\) is the radius of the coil in meters.

### Step 1: Identify the given values
- Number of turns, \(N = 100\)
- Radius, \(R = 2 \times 10^{-3} \, \text{m}\)
- Current, \(I = 1 \, \text{A}\)

### Step 2: Substitute the values into the formula
We can substitute the values into the formula for \(B\):

\[ B = \frac{4\pi \times 10^{-7} \times 100 \times 1}{2 \times (2 \times 10^{-3})} \]

### Step 3: Calculate the denominator
Calculate \(2R\):

\[ 2R = 2 \times (2 \times 10^{-3}) = 4 \times 10^{-3} \, \text{m} \]

### Step 4: Substitute the denominator back into the formula
Now substitute \(2R\) back into the formula:

\[ B = \frac{4\pi \times 10^{-7} \times 100 \times 1}{4 \times 10^{-3}} \]

### Step 5: Simplify the expression
Now simplify the expression:

\[ B = \frac{4\pi \times 10^{-5}}{4 \times 10^{-3}} = \pi \times 10^{-5} \, \text{T} \]

### Step 6: Calculate the value of \(\pi\)
Using \(\pi \approx 3.14\):

\[ B \approx 3.14 \times 10^{-5} \, \text{T} \]

### Step 7: Final answer
Thus, the magnetic field at the center of the coil is:

\[ B \approx 3.14 \times 10^{-5} \, \text{T} = 3.14 \times 10^{-5} \, \text{Tesla} \]